Answer:
Step-by-step explanation:
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A ball with mass m kg is thrown upward with initial velocity 22 m/s from the roof of a building 17 m high. Neglect air resistance. Use g=9.8 m/s2. Round your answers to one decimal place. (a) Find the maximum height above the ground that the ball reaches. xmax= meters (b) Assuming that the ball misses the building on the way down, find the time that it hits the ground.
a) Using the equation of motion formula;
v² = u²+2gH where;
u is the initial velocity
v is the final velocity
theta is the angle of launch
g is the acceleration due to gravity.
H is the maximum height reached by the ball
Since the ball is thrown upwards, the acceleration due to gravity will be negative. The equation then becomes;
v² = u²-2gH
Given
v = 0m/s
u = 22m/s
g = 9.8m.s²
0² = 22²-2(9.8)H
-22² = -19.6H
H = -22²/-19.6
H = 24.69m
If the biuliding is 17m high, the maximum height above the ground that the ball reaches will be;
Hmax = 24.69+17
Hmax = 41.69m
b) The time it takes to hit the ground can be expressed using the formula
v = u-gt
0 = 22-9.8t
-22 = -9.8t
t = -22/9.8
t = 2.45secs